Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$.

Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.

I spend part of the day studying this question Positive integer multiples of an irrational mod 1 are dense and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer.

This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$.

Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.

I think this problem has not been solved.


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    $\begingroup$ a way to get started: you will never hit the same point twice because $x$ is irrational. Therefore you're getting more and more points in the interval $[0,1)$ hence there must be two of them very close to eachother, say closer than any $\epsilon > 0$. see where I'm going with it? $\endgroup$ – mm-aops Jun 22 '14 at 17:12
  • $\begingroup$ I'm not sure. You are using the Weierstrass theorem. But I don't understand what is going on your idea. $\endgroup$ – YTS Jun 22 '14 at 18:00
  • $\begingroup$ assume you have two points $a_n$ and $a_m$ that are closer than $\epsilon$ to each other. can you produce more such pairs of points? $\endgroup$ – mm-aops Jun 22 '14 at 18:14
  • $\begingroup$ math.stackexchange.com/questions/450493/… $\endgroup$ – Patissot Jun 22 '14 at 18:19
  • $\begingroup$ Yes I can. But I don't see your point. By Weiertrass theorem I can get a limit point $x$ and then a subsequence converging to $x$. This subsequence is a Cauchy sequence then there is an infinite number of points such that theirs distance is $<\epsilon$. But I don't see your point :P $\endgroup$ – YTS Jun 22 '14 at 19:03

Ok, since you've asked and it doesn't fit into a comment, there you go. I'll do it on a circle since it's slightly easier to explain and I'll leave it to you to complete it in the case of an interval. let's say you have a circle of length $1$. you take 'steps' along the circle of an irrational length, let's say counter-clockwise. you'll never hit the same spot twice so for any fixed $\epsilon > 0$ you'll eventually find two 'steps' $a_n$ and $a_m$ such that $0 < |a_n - a_m| < \epsilon$. the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$ and so on. therefore if you let $k:= n-m$ and you only consider each $k$-th step you'll be going around the circle travelling a distance smaller than $\epsilon$ hence if you divide your circle into arcs of equal lengths greater than $\epsilon$ (but just slightly, say smaller than $2 \epsilon$) you'll have to land in each one of those in order to make your way all around the circle (because your steps are to small to jump over them). Every point of the circle is in at least one of those intervals which means that for each point of the circle you can find a number $a_j$ in your sequence that is closer than $2 \epsilon$ to it. Now conclude taking smaller and smaller $\epsilon$'s.

edit: oh, just note that I'm taking the distance along the circle, not the euclidean one

  • $\begingroup$ Why the distance beetwen $a_{n-m}$ and $a_0$ is the same as $a_n$ and $a_m$? $\endgroup$ – YTS Jun 23 '14 at 19:00
  • $\begingroup$ because the number of steps separating these points is the same and it doesn't matter if I start from $a_n$ or $0$. you can just rotate the circle so that $a_n$ lands on $0$. (oh, and $0$ means $0$ in the circle, so $1$ in $\mathbb{R}^2$, it's just a way of speaking) $\endgroup$ – mm-aops Jun 23 '14 at 19:05

Note that your set $X$ pretty much takes the decimal values of your irrational number $x$ multplied by $n\in \mathbb{N}$.

Take x=0.01234567890001020304050607080910....

x is such that the first ten digits after the decimal specify the first possible digit of [0,1), the successive $2\times 10 \times 10$ digits would specify any first 2 possible digits of [0,1), the successive $3\times 10 \times 10 \times 10$ would specify any first 3 possible digits of [0,1).

Now taking the x that I proposed, one can attain any value of $r\in [0,1)$ up to the $i$th digit by multiplying x according to some value $10^k\in \mathbb{N}$ (where $k\in \mathbb{N}$).

The details are left to anyone that has to write the proof themselves. In overview, one would take a $\epsilon>0$ and any $r\in [0,1)$ and find out how tiny $\epsilon$ is. Take for example $\epsilon$ has $d$ zeros before non zero digits start, then one would want to specify the first $d+1$ elements of $r$ by multiplying $x$ accordingly.

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    $\begingroup$ How does this work for an arbitrary irrational? You need to prove some properties of irrationals to generalise the above as far as I can see. $\endgroup$ – copper.hat Jun 23 '14 at 2:05
  • $\begingroup$ Yes, you are right. I mis-read the question. I cannot think of a way to generalize this at the moment. $\endgroup$ – Sixter Jun 23 '14 at 2:40

I think I found the solution.

If $x$ is irrational then $\mathbb{Z} + x\mathbb{Z}$ is dense in $\mathbb{R}$ (it's a well known result) and $X = (\mathbb{Z} + x\mathbb{Z}) \cap [0,1[$ because if $k + nx \in [0,1[$, then $k=-\lfloor nx\rfloor$.

  • $\begingroup$ could you edit your answer using meta.math.stackexchange.com/questions/5020/… ? $\endgroup$ – user190080 Feb 1 '17 at 19:15
  • $\begingroup$ Do i need a computer to use it ? $\endgroup$ – user411953 Feb 1 '17 at 19:30
  • $\begingroup$ In some sense. You need some sort of computery device to access the site at all. $\endgroup$ – Daniel Fischer Feb 1 '17 at 19:39
  • $\begingroup$ On another note, I'm not sure that using the fact that the only proper closed subgroups of $\mathbb{R}$ are the discrete ones without proof is appropriate here. And the question is about the denseness of $(\mathbb{Z} + x\mathbb{N}) \cap [0,1[$, not $\mathbb{Z} + x\mathbb{Z}$. It requires further arguments to get the denseness of $(\mathbb{Z} + x\mathbb{N})\cap [0,1[$ from that of $(\mathbb{Z} + x\mathbb{Z})\cap [0,1[$. $\endgroup$ – Daniel Fischer Feb 1 '17 at 19:47
  • $\begingroup$ Its just simple typesetting, you can use whatever device you used to comment or to write your answer in the first place $\endgroup$ – user190080 Feb 1 '17 at 20:02

$f \colon x \rightarrow x-\lfloor x \rfloor$

$F=\{f(nx),n \in \Bbb{Z}\}$

$F \neq \emptyset$ and $x \ge 0$ $\exists d=\inf(F)$

We can prove that $d=0$ (suppose a $\gt$ 0 and ...)

hence $\forall$ y $\gt$ 0 $\exists$ n $\in$ $\Bbb{N}$ , $y \gt f(nx) \gt 0$

$a,b \in [0;1[$ if $b-a \gt 0$ $\exists n \in \Bbb{N}$, $b-a \gt f(nx) \gt 0$

$\exists p \in \Bbb{N}$ , pf(nx) $\gt$ a (1) , if we suppose that p is the smallest number who verifies (1) we have that (p-1)fn(x) $\lt$ a therefore pf(nx) $\lt$ a+f(nx) $\lt$ b

therefore pf(nx) $\in$ [a;b] and pf(nx)=f(pnx)

we can conclude about the density

First time I use this way of writing I hope I have been clear


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